2 different vowels and 2 different consonants are to be selected from the English alphabet. Since there are 5 vowels in the English alphabet, number of ways of selecting 2 different vowels from the alphabet
\(=\space^5C_2=\frac{5!}{2!3!}=10\)
Since there are 21 consonants in the English alphabet, number of ways of selecting 2 different consonants from the alphabet
= \(^{21}C_2\)
= \(\frac{21!}{2!9!}\)
\(= 210\)
Therefore, number of combinations of 2 different vowels and 2 different consonants = 10 × 210 = 2100
Each of these 2100 combinations has 4 letters, which can be arranged among themselves in 4! ways.
Therefore, required number of words = \( 2100 \times 4!\) = 50400